Infrared Spectrometry
- A pair of atoms joined by a chemical bond constantly vibrate. Heavy atoms will cause the bond to vibrate more slowly, and stronger bonds will vibrate more quickly.
- The absorption of infra-red radiation increases the vibrational energy of a bond, by either bending or stretching
- The degree of vibration depends on bond length, bond strength and the mass of the atoms joined
- Particular bonds, and therefore functional groups, can be identified by looking at the frequencies absorbed in a infrared spectrum.
- An infra-red spectrum is produced by a spectrometer:
- A beam of IR radiation in the range 200 – 4000 cm-1 passes through a sample
- The molecule absorbs some of the frequencies, and the emerging beam is analysed to identify the frequencies that have been absorbed
- A computer plots a graph of transmittance against wavenumber
- Each peak in the spectrum represents the absorption by a specific bond vibration in the molecule
- The bonds present in greenhouse gases (carbon dioxide, methane & water) can absorb infrared radiation that is reflected by the Earth’s surface, contributing to global warming.
Mass Spectrometry
- A molecular ion (M+) is formed when molecules in a sample are ionised.
- The molecular ion peak (the peak with the highest m/z ratio in the mass spectrum) can be used to determine the relative molecular mass
- Molecular ions are able to break up into fragments during ionisation. This creates a fragmentation pattern on the mass spectrum. Different compounds will produce different fragment peaks
CH3OH → CH3OH+ + e- → CH3+ + OH•
Molecular ion fragment ion
Molecular ion fragment ion
Combined Techniques
- The wide variety of analytical techniques that we have come across can be used to provide data and information to help us determine the structure of unknown compounds.
- The molecular formula of a compound can be determined using the empirical formula and relative molecular mass of the molecule.
- E.g. Determine the molecular formula of a compound with empirical formula CH2 and a relative molecular mass of 224. Relative molecular mass of the empirical formula:
C H2
12 + (1 × 2) = 14
12 + (1 × 2) = 14
Divide the relative molecular mass by that of the empirical formula:
224/14=16
Molecular formula:
16 x Ch2 = C16H32
Molecular formula:
16 x Ch2 = C16H32
Empirical & Molecular Formula
- The empirical formula is the simplest whole-number ratio of atoms of each element present in a compound.
- The empirical formula can be calculated from the composition by mass or percentage by mass. e.g 6.2 g of P is combined with O2 to form 14.2 g of phosphorous oxide. Calculate the empirical formula of the compound.
mass of O2: 14.2 g - 6.2 g = 8 g
number of moles of each element:
Divide through by the smallest number of moles to get the whole number ratio:
- The molecular formula gives the number and type of atoms of each element in a molecule. It is made up of a whole number of empirical units.
- The molecular formula can be determined using the empirical formula and relative molecular mass of the molecule. e.g. Determine the molecular formula of a compound with empirical formula CH2 and a relative molecular mass of 224.
Relative molecular mass of the empirical formula:
Empirical formula: P2O5
C H2
12 + (1 × 2) = 14
Divide the relative molecular mass by that of the empirical formula:
224/14=16
Molecular formula:
16 x CH2 = C16H32
12 + (1 × 2) = 14
Divide the relative molecular mass by that of the empirical formula:
224/14=16
Molecular formula:
16 x CH2 = C16H32